course Mth 158 9/25 4:15 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is • (x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to • (x+2)/[(x-2)(x^2+4)]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I know it takes me the long way around to get the answer, but for now it’s the only way I can do it. I keep reading the chapters over and over and each time I understand a little more. Hopefully, it will just come to me and I will be able to work the problem as easily as the given solution.
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Given Solution: [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2x – 5) / (3x + 2) + (x + 4) / (3x + 2) Since we have the same value as both denominators it would be written (2x – 5) + (x + 4) / (3x + 2) = (3x – 1) / (3x + 2) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We have two like terms so we write • (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have • (3x-1)/(3x+2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * R.7.52 (was R.7.48). Show how you found and simplified the expression (x - 1) / x^3 + x / (x^2 + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-1) / x^3 + x / (x^2 + 1) In this case you would multiply the first set by the x^2 + 1 / x^2 + 1 I don’t know how to set this up with typewriter notation, but I’ll try (x-1) * (x^2 + 1) / (x^3) * (x^2 + 1) and then add that to (x) * (x^3) / (x^2 + 1) * (x^3) Reducing it to (x-1)(x^2 +1) / (x^3)(x^2 +1) + x^4 / (x^3)(x^2 +1) x^3 + x – x^2 – 1 + x^4 / (x^3)(x^2 + 1)
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Given Solution: Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / / [ (x^3)(x^2+1)] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t simplify the last step. I worked this on paper last night and then copied it over to word so I could post it, but when I was re-typing it, I couldn’t remember how I got the x^3 + x – x^2 – 1 + x^4. I am sure it is a formula, but I don’t remember which one I used. Can you tell me how to recognize what formula I need to use to solve a problem? I think that is where I’m spending most of my time trying to figure out what to do and looking it up in the text book. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the LCM, you find everything that factors into each one of the problems. I look at it like a universal set and then the rest are subsets of that set. x – 3 is just x – 3 x^3 + 3x factors to x ( x^2 + 3) x^3 – 9x factors to x (x-3)(x+3) Then I find each sub set and put them together to make a universal set LCM = x (x-3) (x^2 +3) (x + 3) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: • LCM = x(x-3)(x+3)(x^2+3) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t know if my answer is acceptable in the way that I find the answer, but it is the only way I can understand how to solve it at the moment.
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Given Solution: * * ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2. To express the given expression in terms of the common denominator we then multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t see where the 3x(x-1) comes in. When I tried working it on paper, I started with 3x / (x -1) – (x-4) / (x^2 - 2x + 1) and reduced to 3x / (x-1) – (x-4) / (x-1)^2 But then I tried to solve it by (3x) – (x-4) / x-1 which isn’t right